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WARRIOR [948]
3 years ago
15

The spring of a spring balance is 5.0in. long when there is no weight on the balance, and it is 8.4in. long with 4.0 lb hung fro

m the balance. How much work is done in stretching it from 5.0in. to a length of 12.8in.?
Mathematics
1 answer:
djyliett [7]3 years ago
4 0

Answer: work is done in stretching it from 5.0in. to a length of 12.8in is 36 lb.In

Step-by-step explanation:

We know that

f(x) = kx

given that;

Natural length L = 5

x = 8.4 - 5 = 3.4

force F = 4.0 lb

f(x) = kx

we substitute

4 = k × 3.4

k = 4 / 3.4

k = 1.1764 = 20/17

Now work done from natural length to 12.8

x = 12.8 - 5 = 7.8

work done = ₀∫^7.8  f(x) dx

= ₀∫^7.8  kx dx

we substitute value of k

= ₀∫^7.8  (20/17)x dx

= 20/17 [ x²/2 ]

= 20/17 [ (7.8)²/2 - 0 ]

= 1.1764 × 30.42

= 35.786 ≈ 36 lb.In

Therefore work is done in stretching it from 5.0in. to a length of 12.8in is 36 lb.In

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Step-by-step explanation:

Given:

Diameter of each ball (d) = 13 cm

Radius of the ball (r) = radius of the cylinder = ½(13) = 6.5 cm

Height of the cylinder (h) = the sum of the diameters of the 3 balls = 3(13) = 39 cm

a. Volume of cylinder = πr²h

Plug in the values

Volume of cylinder = π*6.5²*39 = 5176.56 ≈ 5177 cm³

b. Total volume of the three balls = 3(volume of 1 ball) = 3(volume of sphere) = 3(⁴/3πr³)

Plug in the value

Total volume of the 3 balls = 3(⁴/3*π*6.5³) = 3(1150.35) = 3,451.05 ≈ 3,451 cm³

c. % of volume of the container occupied by the 3 balls = total volume of the three balls / volume of cylinder × 100

Plug in the values

= 3,451/5177 × 100

= 66.6602279

≈ 67%

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